Re: The Emission Theory of Androcles

"Inertial" <relatively@xxxxxxxx> wrote:
"Jonah Thomas" <jethomas5@xxxxxxxxx> wrote
"Inertial" <relatively@xxxxxxxx> wrote:
"Jonah Thomas" <jethomas5@xxxxxxxxx> wrote

The alternative way that
Inertial and I were thinking went more like this:

S /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
D S/\/\/\/\/\/\/\/\/\/\/\

I don't follow you .. D is behind S there

Yes, because the emitter is moving with the detector. By the time
the wavefront which was emitted at D reaches the detector the source
has moved behind D on the forward side and ahead of D on the back
side. That's why the distances are different.

No .. the emitter (as you said) moves with the detector. The path
from emitter to detector is the SAME LENGTH all the time !!

Yes, but the path that light must take to get from the emitter to the
detector is different in the two directions.

Because the detector moves away from one light beam and moves toward the
other one. We all agree on that, right?

In the case of the fast light, it's fast because the emitter is moving
with it. But the detector is moving away from it too. So in my
animation, the path is about 10% longer than it would be if both paths
were the same length. And the slow light has a path that's about 10%
shorter than it would be if both were the same length. That's correct,
isn't it?

I show the fast light moving back a bit and the slow light moving
forward because that to me fit the detector moving away from the fast
light and toward the slow light. But I could draw that some other way if
you preferred. I wanted both of them to end up in the same place so it
would be completely obvious whether they were in phase at the end.

I think you meant at the start (just before rays emitted) we have:

S D
S D

One third of the way thru we have this for the two rays, where s is
the stationary point s we marked in the non-rotating frame, and R
is> the leading edge of the ray (ie the photon/wave/whatever that was
first emitted):

s S/\/\/\/R D
S/s/\/\/R D

You can S has been making more photons/waves/whatevers since R ..
they> come from S's current position, not from s!!

Yes, exactly.

Two thirds of the way through we have even more
photons/waves/whatevers from S:

s S/\/\/\/\/\/\/\/R D
S/\/s/\/\/\/\/\/R D


At the end we have
s S/\/\/\/\/\/\/\/\/\/\/\/D
S/\/\/s/\/\/\/\/\/\/\/\/D

The rays arrive at D in phase, they are still in phase at the
source S> as well.

What happens at s doesn't make any difference !!!!

They are in phase at D in the model you use, because you assumed
they would stay in phase and required them to do so.

They can't be anything BUT be in phase in a non-relativistic situation
with constant speeds and same arrival time.

That's because you assume that the front of the wave stays the same for
both.

And that's a reasonable assumption. If you figure that when the light
left the emitter it started at zero and the electric field then climbed
to positive one and then fell to zero, why would the same spot on the
wave later be anything but zero trending positive?

Still, that is an assumption. And when I drew the picture your way, with
the wave reaching the detector in phase and then I drew it backward to
the sources with constant wavelength and speed 0.9 and 1.1 and distance
11 and 13, it was out of phase at the source. That does not work.
Agreed? Surely you don't have one wave leave the emitter at some plus
value at the same time the other leaves with some minus value.

When I draw it in phase at the source, and then I draw the waves with
the same constant wavelength, I get it out of phase at the detector.

Did I choose some parameter wrong? I tested some things. Yes, the
wavelength is the same. One of them goes 1.1 while the other goes 0.9.
What did I do wrong? Would you like my source code, or would you want to
do it yourself from scratch and see what you get?

Unless something somehow changes the frequency relative to the
detector .. but if the frequency is the same and arrival time the same
(all relative to the detector) then it MUST be in phase.

Why would the frequency be the same? The wavelength is the same and the
speed is different. I'd sure expect the frequency to be different.

In the model Wilson uses, they do not need to be in phase.

HOW !!!!!! .. if they are emitted in phase they arrive in phase.
Nothing happens in between to change that

Check your assumptions? He isn't using the same assumptions you are. Are
his wrong? How can you find out if you're so stuck on your own that you
can't even understand his?

His model
works just fine. Unless I made an arithmetic mistake his model
works, it gets a phase shift. I haven't checked whether it is the
right phase shift.

There is no phase shift

My picture could be wrong. Maybe I made a mistake that I haven't found.
If you look at it closely you'll see that it has little bits of wave
evaporating off the front of one wave, and extra little bits magically
appearing at the front of the other wave. There's no reason you'd
predict that would happen, right? But it has to happen if the rest of it
is right. The wavelength is the same, and the frequency at the source is
the same, and the wave is traveling faster. So extra wave appears at the
front. Is that possible? Well, you certainly wouldn't have expected it,
right? Is there maybe something in Maxwells Equations that says it's
impossible?

If you want to argue about whether his assumptions are unreasonable
we can, but I don't think it's arguable whether he gets a phase
shift or not.

Yes .. it most definitely is worth arguing about.

His has a fixed source in the inertial frame and a moving detector,
with the fixed source emitting two different frequency waves at two
different speeds.

It's a moving source, right?

This is nothing like Sagnac

I drew the picture straightened out because that was easiest for me. Was
that the problem?

http://i847.photobucket.com/albums/ab31/jehomas/speedwave4.gif
http://i847.photobucket.com/albums/ab31/jehomas/speedwave6.gif

The points on the left side that move left and right (and so get
further and closer to the destination) correspond to where the source
WAS when the FIRST photon/wave/ray was emitted. But you show waves
continually emitted from that location. That is NOT the case.

No, they correspond to where the source is when it's emitting more. The
original starting source is elsewhere.

The photons/waves/rays are emitted a constant distance from D at all
times.

Yes. Would it help if I label D and the original S? Why do I ask, of
course it would.

You are showing the waves (in the rotating frame point of view)
continually emitted a source point that changes distance from D (ie
our point s above). There is NO SUCH SOURCE OF WAVES IN SAGNAC !!!!

?? In Sagnac, when the detector moves the source moves with it,
right?

Yes .. you don't show that .. you have the distance from source to
detector changing !!!

I didn't put the detector in, the detector was where they stop. Clearly
I need to label it better.

So one source appears to get farther from D while the other appears
to get closer.

No .. can't you read that you just contrradicted yourself "detector
moves with source" "source..get farther..get closer"

Pick a time when a wave leaves the source in one direction, and another
leaves the opposite direction. The detector is moving away fron one wave
and toward the other. So later waves will start from farther behind on
one side and closer on the other.

The source CANNOT get farther and closer to the detector when ther is
ONE source and it moves WITH the detector

Please try to bear with me, I'm not that good at saying this stuff yet.
It's new to me.

Because by the time the light that is getting emitted later
reaches D, D will be farther away.

No !!!!! The detector is always THE SAME DISTANCE FORM THE SOURCE.

It would be easier to show it with a picture, but my current asnimation
software is a little limited for that. I'll extend it or find something
better, but this is a spare-time project....

Once we assume constant wavelength, it is absurd to have the
waves> > get out of phase at the actual source. Wilson's alternative
is the> > only one that can make sense,

No .. it doesn't

It might not, but as I understand it yours makes no sense at all
with constant wavelength.

It works perfectly with fixed wavelength

No, when I make it fixed wavelength and make them be in phase at the
detector, they are out of phase at the source. I don't consider that
"works perfectly". You can do it if the wavelength isn't fixed.

You can argue that it can't be constant wavelength,

I'm not

or you can show me how I misunderstood your constant-wavelength
model.

I don't know what your misunderstanding could possibly be, so I can't
correct it

How does your constant-wavelength model work?

1. Fixed distance from D to S.

2. Rotates at 0.1c

3. Light in one direction travels at 0.9c, light in the other direction
travels at 1.1c.

4. The distance traveled is 1.1 times the total fixed distance in one
case, and 0.9 times the fixed distance in the other case.

5. Both directions have the same fixed wavelength, the same wavelength
they have with no rotation, when both directions travel at c and the
distances are the same.

6. Both light beams are in phase when they leave the source and in phase
when they arrive at the detector.

Did I leave something out? Did I add something extra? I can only make
this work for certain particular distances that just happen to match up
the lengths to be a whole number of wavelengths.

unless we find a way to change hidden assumptions I
didn't notice I was making.

If Wilson's approach doesn't work either then it will probably
turn> > out that it simply does not make sense to have waves with
constant> > wavelength in this circumstance.

Why is all this so hard for you and Henry?

Wilson thinks he knows something you don't,

BAHAHAH

but he has trouble explaining it.

Because he doesn't .. he's just making stuff up and confusing himself
(and you) and doesn't know what-the-hell he's talking about.

I'm trying to understand what he's saying and whether he
can be right.

If he had a consistent model for the scenario, and if his scenario was
the same as the Sagnac we are discussing, then that would be a
worthwhile pursuit.

If he doesn't know what he's talking about, then he's challenging me to
make sense of his cryptic words. I'm starting to do so. I like that
almost as much as if he understands it too.

My previous objection was wrong. I thought I was agreeing
with you, and if so you were wrong too. But I may have misunderstood
you just like I misunderstood him for so long.

Quite likely.

Finding that the reason I thought he was wrong was itself wrong
doesn't make him right. But it leaves open the possibility. I don't
know whether I'll find other objections. And of course, his model
can work and still not fit the reality.

No .. a model that doesn't fit reality isn't a working model

We'd have to find out how it doesn't fit reality. It fits the Sagnac
experiment better than Pauli thought it did, unless I've made a mistake.

But I was wrong to say that he couldn't have his
Sagnac waves get out of phase.

He can't !!!

If you think he can .. please explain how it can happen.

I hope you understood my explanation. I'll try to draw the picture
better.

The source of the waves/photons/whatever MOVES WITH THE DETECTOR.

Yes.

The distance between the source and the detector IS THE SAME ALL THE
TIME.

Yes.

The distance between where the source WAS and the detector is NOW
changes, because the source moves.

Yes.

if the waves/photons/rays are in phase at the source, and have the
same speed and frequency relative to the source (they do) and if when
they arrive at they arrive at the detector at the same time, with the
same speed and frequency relative to the detector (they do) then they
must still be in phase.

Ah. There is your hidden assumption. You assume that the wavelength
times the frequency must equal the speed.

Once you give up that assumption then they can get out of phase.

The source is moving, so the frequency at the source does not have to
equal the frequency of the wave movement.

So anyway, I'll redraw the pictures and make them clearer, and maybe
I'll find a mistake in the process.
.

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